A student prepares a solution of hydrochloric acid that is approximately 0.1 M and wanted to determine its precise concentration. A 25.00 mL portion of the HCl solution is transferred to a flask, and after a few drops of indicator are added, the HCl ?

1 Answer
Stefan V.
Mar 28, 2015

The molarity of the hydrochloric acid solution is #"0.095 M"#.

You're performing a titration of the #"HCl"# solution using sodium hydroxide. The balanced chemical equation for the neutralization reaction that takes place is

#HCl_((aq)) + NaOH_((aq)) -> NaCl_((aq)) + H_2O_((l))#

Notice the #"1:1"# mole ratio that exists between sodium hydroxide and hydrochloric acid - at the equivalence point, the number of moles of the former will be equal to the number of moles of the latter.

The number of moles of added sodium hydroxide is

#C = n/V => n = C * V#

#n_("NaOH") = 38.71 * 10^(-3)"L" * "0.0615 M" = "0.002381 moles"# #"NaOH"#

The number of moles of hydrochloric acid will be equal to

#n_("HCl") = n_("NaOH") = "0.002381"#

This means that the molarity of the hydrochloric acid solution will be

#C = n/V = "0.002381 moles"/(25.00 * 10^(-3)"L") = "0.095 M"#

Related questions
See all questions in Titration Calculations
Impact of this question
4027 views around the world
You can reuse this answer
Creative Commons License